Question

If $\alpha$ and $\beta$ are the roots of the equation $x^2-4\sqrt{2}kx+2e^{4lnk}-1=0$ for some $k$ and ${\alpha }^2+{\beta }^2=66$, then ${\alpha }^2+{\beta }^2+\alpha +\beta$ is equal to

• A. $66+8\sqrt{2}$
• B. $74$
• C. $66-8\sqrt{2}$
• D. $58$

Answer 1

The correct option is A $66+8\sqrt{2}$

We have,
$\alpha +\beta =4\sqrt{2}K$
$\alpha \beta =2k^4-1$
${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta$
$66=16\times 2k^2-4k^4+2$.
$\Rightarrow k^4-8k^2+16=0$
$\Rightarrow (k^2-4{)}^2=0$
$\Rightarrow k^2=4$
$k=\pm 2$, since $k>0,k=+2$
So, equation becomes $x^2-8\sqrt{2}x+31=0$
${\alpha }^2+{\beta }^2+\alpha +\beta$
$=66+4k\sqrt{2}$
$=66+8\sqrt{2}$